It was already mentioned somewhat, but you need some sort of weight metric for 
each job.  Save the “smallest” job.  For each driver, assign the largest and 
smallest job remaining.  When done, add the weight for each driver and give the 
remaining job to the one with the least work.

Thanks,
Brian
On Apr 16, 2020, 10:34 PM -0400, Skip Kimpel via use-livecode 
<use-livecode@lists.runrev.com>, wrote:
> Haha... Let me specify, I am looking to do this programmatically :)
>
> SKIP
>
> On Thu, Apr 16, 2020 at 10:31 PM Dev via use-livecode <
> use-livecode@lists.runrev.com> wrote:
>
> > Since you would probably have long trips and short trips, I would try and
> > allocate on an estimated time to complete basis so that all were busy for
> > about the same time.
> >
> > > On 16-Apr-2020, at 8:23 PM, Skip Kimpel via use-livecode <
> > use-livecode@lists.runrev.com> wrote:
> > >
> > > I have 15 "jobs" that need to be assigned to 7 "drivers". How would you
> > > evenly distribute the jobs to the drivers? In this case one of the
> > drivers
> > > would have 3 jobs while the others have two.
> > >
> > > Obviously, the number of drivers and number of jobs would fluctuate.
> > >
> > > This has been my mind twister for the night thus far :)
> > >
> > > SKIP
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